What does a morphism in the category $\mathsf{C}_{\alpha, \beta}$ correspond to?

Question

Start with a given category $\mathsf{C}$ and fix two morphisms $\alpha: A \to C$ and $\beta: B \to C$. Consider the category $\mathsf{C}_{\alpha, \beta}$ as follows:

---

Does a morphism in this category correspond to:

$1.$ a morphism $\sigma: Z_1 \to Z_2$ such that $f_1=f_2\sigma$ and $g_1=g_2\sigma$?

$2.$ a morphism $\sigma: Z_1 \to Z_2$ such that $\alpha f_1=\alpha f_2\sigma$ and $\beta g_1=\beta g_2 \sigma$?

$3.$ a morphism $\sigma: Z_1 \to Z_2$ such that $\alpha f_1=\beta g_2\sigma$ and $\beta g_1=\alpha f_2 \sigma$?

It seems that if (1.) is true, then the others are also true.

Answer 1

In order for the diagram to be commutative, all of its sub-diagrams must commute. This means, in particular, that the equations in condition (1) must hold. As you suggest, (2) and (3) follow from (1).

So a morphism $\sigma : (f_1,g_1) \to (f_2,g_2)$ in $\mathsf{C}_{\alpha,\beta}$ is a morphism $\sigma : Z_1 \to Z_2$ in $\mathsf{C}$ such that $f_2 \circ \sigma = f_1$ and $g_2 \circ \sigma = g_1$.

Answer 2

A morphism in $\mathcal{C}_{\alpha,\beta}$ between objects $(Z_1,f_1,g_1)$ and $(Z_2,f_2,g_2)$ is a morphism in $\mathcal{C}$ between $Z_1$ and $Z_2$ such that the diagram that you showed commutes.

Yes, it is true that if $(1)$ holds, then $(2)$ and $(3)$ also hold. However, in general, it is not true that either $(2)$ or $(3)$ imply $(1)$ (i.e., you can meet $(2)$ or $(3)$ and not $(1)$), so the conditions are not equivalent.

Furthermore, you can easily show that the diagram is equivalent to $(1)$: just looking at the left part of the diagram, you easily see that it implies $(1)$. But in fact, they're equivalent because the rest of the diagram just follows from $Z_2$ being an object.

So $(1)$ is the only correct answer, as the others are too weak (i.e., they hold more often than the diagram does).

I hope this helps.